\section{Bandwidth Calculation OOK modulation}
\label{sec:bandwidth}
The Modified Miller encoded data for the forward channel results into a signal with off pulses with a length of 3μs. The 13.56MHz signal is then OOK modulated with the data signal. The minimum bandwidth required to read out this modulated signal can be found by looking at the Fourier transformation of the pulses.

The Fourier transform of the encoded data is:

\begin{equation}
\mathcal{F}\{1-rect(\frac{t}{3\cdot10^6})\}=\delta(\omega)-3\cdot10^6\cdot sinc(\pi\cdot3\cdot10^6\cdot\omega)
\end{equation}
The signal in time domain and frequency domain is shown in figure \ref{fig:datapulse}.

\begin{figure}[h]
\includegraphics[width=\textwidth]{assets/pulseandfourier.png}
\caption{Data pulse in time domain (left) and frequency domain (right)}
\label{fig:datapulse}
\end{figure}

The right plot of figure \ref{fig:datapulse} shows that most of the power of the signal lies between the middle two zero crossings at $\pm$333kHz. A bandwidth of $\pm$333kHz would therefore be sufficient to allow the data signal to pass.